- #1
Utilite
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Okay, I am studying Baby Rudin and I am in a lot of trouble.
I want to show that a closed interval [a,b] is compact in R. The book gives a proof for R^n but I am trying a different proof like thing.
Since a is in some open set of an infinite open cover, the interval [a,a+r_1) is in that open set for an r_1. Similarly a+r_1 is in the closed interval, [a+r_1,a+r_1+r_2) is in an open set. Going like this we can reach an r_k such that a+r_1+r_2...+r_k is greater than or equal to b. Therefore an interval [c,b] is an open set and those open sets form a subcover of an infinite open cover. M=min{r_1,r_2...r_k}>0 Therefore k needs to be less than or equal to (b-a)/M which is clearly finite. Even if every interval [a+r_1...r_i,a+r_1...r_(i+1)) is contained in different open sets the number of open sets is finite.
Every infinite open cover of an interval [a,b] has a finite subcover then it is compact.
I think I am making a mistake because this should work for [a,b). But [0,1) is not compact.
I am very confused, please help me out.
I want to show that a closed interval [a,b] is compact in R. The book gives a proof for R^n but I am trying a different proof like thing.
Since a is in some open set of an infinite open cover, the interval [a,a+r_1) is in that open set for an r_1. Similarly a+r_1 is in the closed interval, [a+r_1,a+r_1+r_2) is in an open set. Going like this we can reach an r_k such that a+r_1+r_2...+r_k is greater than or equal to b. Therefore an interval [c,b] is an open set and those open sets form a subcover of an infinite open cover. M=min{r_1,r_2...r_k}>0 Therefore k needs to be less than or equal to (b-a)/M which is clearly finite. Even if every interval [a+r_1...r_i,a+r_1...r_(i+1)) is contained in different open sets the number of open sets is finite.
Every infinite open cover of an interval [a,b] has a finite subcover then it is compact.
I think I am making a mistake because this should work for [a,b). But [0,1) is not compact.
I am very confused, please help me out.